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Question Number 121502 by bramlexs22 last updated on 08/Nov/20

 (1) lim_(x→0)  ((√(1−cos 2x))/(x(√2))) ?  (2) ∫ cot^3 x csc^3 x dx

(1)limx01cos2xx2?(2)cot3xcsc3xdx

Commented by liberty last updated on 09/Nov/20

(2) ∫ cot^2 x csc^2  x(cot x csc x )dx =  ⇔ ∫ csc^2 x (csc^2 x−1) (cot x csc x)dx=  ⇔ ∫csc^2 x (1−csc^2 x) d(csc x) =  ⇔ ∫( z^2 −z^4 )dz = (1/3)z^3 −(1/5)z^5  + c   ⇔ (1/3)csc^3 z −(1/5)csc^5 z + c   ⇔ (1/(3sin^3 x)) − (1/(5sin^5 x)) + c

(2)cot2xcsc2x(cotxcscx)dx=csc2x(csc2x1)(cotxcscx)dx=csc2x(1csc2x)d(cscx)=(z2z4)dz=13z315z5+c13csc3z15csc5z+c13sin3x15sin5x+c

Commented by liberty last updated on 09/Nov/20

(1) lim_(x→0)  ((√(1−cos 2x))/(x(√2))) = lim_(x→0)  (((√2) (√(sin^2 x)))/(x(√2)))   lim_(x→0)  ((∣sin x∣)/x) → { ((lim_(x→0^+ )  ((∣sin x∣)/x) = lim_(x→0^+ )  ((sin x)/x)=1)),((lim_(x→0^− )  ((∣sin x∣)/x) = lim_(x→0^− )  ((−sin x)/x)=−1)) :}  then lim_(x→0)  ((√(1−cos 2x))/(x(√2))) doesnot exsist

(1)limx01cos2xx2=limx02sin2xx2limx0sinxx{limx0+sinxx=limx0+sinxx=1limx0sinxx=limx0sinxx=1thenlimx01cos2xx2doesnotexsist

Answered by Lordose last updated on 08/Nov/20

1. lim_(x→0) ((√(cos^2 x+sin^2 x−cos^2 x+sin^2 x))/(x(√2)))  lim_(x→0) (((√2)sinx)/(x(√2))) = lim_(x→0) ((sinx)/x) = 1

1.limx0cos2x+sin2xcos2x+sin2xx2limx02sinxx2=limx0sinxx=1

Commented by liberty last updated on 12/Nov/20

wrong

wrong

Answered by Lordose last updated on 08/Nov/20

2. ∫cot(x)csc^3 x(csc^2 x−1)dx   {cot^2 x=csc^2 x−1}  ∫(u^2 −u^4 )du  {u=csc(x) ⇒ du=−ucotxdx}  (u^3 /3) − (u^5 /5) + C  (1/(15))(5csc^3 x − 3csc^5 x) + C

2.cot(x)csc3x(csc2x1)dx{cot2x=csc2x1}(u2u4)du{u=csc(x)du=ucotxdx}u33u55+C115(5csc3x3csc5x)+C

Answered by Bird last updated on 09/Nov/20

we have cos(2x)∼1−2x^2 ⇒  1−cos(2x)∼2x^(2 )  ⇒(√(1−cos(2x)))∼(√2)∣x∣  ⇒((√(1−cos(2x)))/(x(√2)))∼(((√2)∣x∣)/(x(√2)))=f(x) ⇒  lim_(x→0^+ )  f(x) =1 and      lim_(x→0^− )   f(x)=−1

wehavecos(2x)12x21cos(2x)2x21cos(2x)2x1cos(2x)x22xx2=f(x)limx0+f(x)=1andlimx0f(x)=1

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